Question

Given that two of the zeroes of the cubic polynomial $a{x}^3+b{x}^2+cx+d$are 0 , the third zero is
(a) $-\frac{b}{a}$ (b) $\frac{b}{a}$ (c) $\frac{c}{a}$ (d) $-\frac{d}{a}$

Answer 1

Let the polynomial be $f\left(x\right)=a{x}^3+b{x}^2+cx+d$.
Suppose the two zeroes of f (x) are $\alpha =0\mathrm{and}\beta =0$.
We know that,
Sum of the zeros,
$$\begin{gathered} \alpha +\beta +\gamma =\frac{-b}{a} \\ \Rightarrow 0+0+\gamma =\frac{-b}{a} \\ \Rightarrow \gamma =\frac{-b}{a} \end{gathered}$$
Hence, the correct answer is option A.